Three masses each of mass m are placed at the vertices of an equilateral triangle ABC of side l as shown in the figure. The force acting on a mass 2m placed at the centroid O of the triangle is:

The correct option is A zero
Draw a perpendicular AD to the side BC.
$\therefore AD=AB\sin 60=\frac{\sqrt{3}}{2}l$
Distance AO of the centroid O from A is $\frac{2}{3}$AD.
$\therefore AO=\frac{2}{3}\left(\frac{\sqrt{3}}{2}l\right)=\frac{l}{\sqrt{3}}$
By symmetry, $AO=BO=CO=\frac{l}{\sqrt{3}}$
Force on mass 2m at O due to mass m at A is
$F_{OA}=\frac{Gm\left(2m\right)}{{\left(l/\sqrt{3}\right)}^2}=\frac{{6Gm}^2}{l^2}$ along OA
Force on mass 2m at O due to mass m at B is
$F_{OB}=\frac{Gm\left(2m\right)}{{\left(l/\sqrt{3}\right)}^2}=\frac{{6Gm}^2}{l^2}$ along OB
Force on mass 2m at O due to mass m at C is
$F_{OC}=\frac{Gm\left(2m\right)}{{\left(l/\sqrt{3}\right)}^2}=\frac{{6Gm}^2}{l^2}$ along OC
Draw a line PQ parallel to BC passing through O. Then $\angle BOP=30=\angle COQ$
Resolving ${\overline{F}}_{OB}$ and ${\overline{F}}_{OC}$ into two components.
Components acting along OP and OQ are equal in magnitude and opposite in direction. So, they will cancel out while the components acting along OD will add up.
$\therefore$ The resultant force on the mass 2m at O is
$F_R=F_{OA}-(F_{OB}\sin 30+F_{OC}\sin 30)$
$=\frac{{6Gm}^2}{l^2}-(\frac{{6Gm}^2}{l^2}\times \frac{1}{2}+\frac{{6Gm}^2}{l^2}\times \frac{1}{2})=0$